Question

A common tangent of the two circles is

• A. $x+\sqrt{3}y=4$
• B. $x=4$
• C. $x+2\sqrt{2}y=6$
• D. $y=2$

Answer 1

The correct option is C $x+2\sqrt{2}y=6$

Let m be slope of the common tangent

$y=mx\pm 2\sqrt{1+m^2}....\left(i\right)$

$y=m(x-3)\pm \sqrt{1+m^2}...\left(ii\right)$

(i) and (ii) are identical

$\frac{1}{1}=\frac{m}{m}=\frac{\pm 2\sqrt{1+m^2}}{-3m\pm \sqrt{1+m^2}}$

$-3m\pm \sqrt{1+m^2}=\pm \sqrt{1+m^2}$

$\Rightarrow -3m\pm \sqrt{1+m^2}$

$\Rightarrow m=\pm \frac{1}{2\sqrt{2}}$ or one tangent is vertical.

For $m=-\frac{1}{2\sqrt{2}}$

$y=\frac{1}{2\sqrt{2}}\pm 2\sqrt{1+\frac{1}{8}}$

$2\sqrt{2}y+x=6$